real analysis review

8/10/2019 real analysis review

1/5

Review 7/30

Q: Regarding HW7#4: Every Cauchy sequence with a convergent subsequence is itself convergent.

Is this fact used for specific Cauchy sequences? Or to prove general theorems?

A: Yes, to both. Dont have good concrete example of first, but for second, you use it to prove that every[sequentially] compact space is Complete.

Q: If you have a projection P: MNM where Mis closed and N=R, i.e.P: [0, 1]R [0, 1], p(x, y ) = x. What is ppre of the codomain?A: ppre([0, 1])=[0, 1]R. [0, 1] is closed, and ppre([0, 1])=[0, 1]R is also closed, reason: p is continuous,and if you draw it out then its apparent why its closed.

In general, p : MNM, p(m, n) = m, then ppre(U) = UN.

Q: Rolles Theorem proof.

A: We state Rolles Theorem as follows. If f: [a, b] R is continuous and differentiable on (a, b), then iff(a) =f(b), there exists c (a, b) s.t. f0(c) = f b fa

ba .

Substep.Iff: [a, b]R is continuous on[a, b]and differentiable on(a, b)and has a local max/min atc (a, b),then f0(c) = 0.

Pf of Substep. Let c be a local max. (for local min case, replace f withf) (???)So, there is some > 0, with a 6 c < c < c + 6 b,and for which f(c)> f(x) for each x (c , c + ).We wish to find f0(c) = limxc

fx fc

x c. We break this up into two cases: x c and x c+.

Case 1. x

c.

Consider limxcf(x) f(c)

x c . Note f(x)6 f(c), hence f(x) f(c)60. Since x approachesc from the left, we

have that x c < 0. Hence, the quotient fx fcx c

>0, hence f0(c) = limxcf x fc

x c >0.

Case 2. x c+.Consider limxc+

fx fc

x c . Note that f(x)> f(c), hence f(x) f(c)60. Sincex approachesc from the right,

we have that x c > 0. Hence the quotient fx fcx c

6 0, hence f0(c) = limxc+fx fc

x c > 0.

Hence, f0(c) = 0. X

Pf of Rolles Theorem.Let f: [a, b]R is continuous, and [a, b]is compact, by EVT we have that fattainsits max and min on [a, b].

If both min and max occur at f(a) =f(b), then fis constant, and f0(c) = 0 for any c (a, b).If not, i.e. at least one of min and max occur somewhere other than x = aandx = b, then that global extremumis also a local extremum, so by substep f0(c) = 0 where c is where that global extremum is. X

Q: Give concrete example of a space Mwith finitely many points, and what do open covers of that spacelook like?

A: Let M= {1, 2, 4}R.What do open sets in this space look like?

1


2/5

An open set in Mhave to be a subset of M. There are only 8 subsets of M, namely

, {1}, {2}, {4}, {1, 2}, {1, 4}, {2, 4}, {1, 2, 4}.Note all of these sets are open inM. e.g.{1}is open because B1, 1

2

= {1} {1}.

An open cover of M is a set of subsets of Mwhose union is M.

Hence, something likeU= {{1, 4}, {1, 2}, {2},} is an open cover of M.OrU= {{1}, {2}, {4}}, orU= {{1, 2, 4}}, orU=(M).NOT an open cover of M is{, {1}, {2}, {1, 2}}, since 4 isnt it.So, why is M covering compact? Any open cover must be a subset of (M). There are eight choices forelements of an open cover, so at most your open cover will have 8 sets.

This shows that every open cover is already finite.

Hence, any open cover has a finite subcover, namely itself. X

Q: What about closed covers?

A: For any metric space M,{{m} : mM} is a closed cover of M.For eachm M,{m}is closed in M, andS

mM{m}= M.But{{m} : mM} has no subcover other than itself.

Every closed cover has a finite subcover means M has only finitely many points.

Q: Why do open covers say something useful?

A: Open sets are kind of already big. So, if you only need finitely many open sets to cover your space, yourspace isnt that big.

Q: What is the intersection of infinitely many open sets look like?

A: Could get closed set:T

n=1 1

n, 1

n

= {0}.

Could get empty set:T

n=1

(n, ) =.Could get sets neither open nor closed:

Tn=1 1, 1

n

= (1, 0].

Q: HW6#3b. Pick any p~ R2 and any r (0,). Let U ={q~ R2 :kp~ q~k < r}, and let K ={q~R2 : kp~ q~k6 r}. Show that U=K.A: We show that U

Kand K

U.

For U K, just need to show Kis a closed set containing U.UKis easy.To show that Kis closed, one way to do it is to refer to an earlier homework problem, to show that for any

pM, r > 0, Many metric space,{qM: d(p, q)6 r} is closed.Another way to do it is to note that of p~= (p1, p2)then

K= {(x, y )R2 : (xp1)2 + (yp2)26 r2}.

2


3/5

So, if you define f:R2R as f(x, y) = (xp1)2 + (yp2)2, then notice that

K= fpre([0, r2]) =fpre((, r2]){z }eitheroneoftheseworks

.

fis a polynomial, hence continuous. Note K is closed, since [0, r2] and (

, r2)are closed in R.

Now, we show that KU. Basically, need to show that for each point in K is the limit of some convergentsequence in R2 whose terms are all in U.

i.e. let p~=q1~,

let q2~ = pt that is 1/2 between p~ and q~

let q3~= pt that is 2/3 between p~ and q~

let qn~=pt that is

n 1

n of way from p~ to q~.

Let q~K.nN, let qn~ = n 1n q~+ 1

np~. i.e. p~{z}

startatp~

+n 1

n (q~p~){z }

goin directionofq~ fromp~{z }butonlygo

n1

n

of theway

=n 1

n q~+

1

np~.

Thenkqn~p~k= n 1

n (q~p~)= n 1

n kq~p~k6 n 1

n r < 1r = r.

So, qn~U for all n N.Notekqn~ q~k=

1n

q~+ 1

np~= 1

nkq~p~k.

Let > 0, if N=r

, then for any n

N with n > N,

kqn~ q~k=1nkq~p~k6 r

n x}, K{(x, y) :y >x}

show K= U.

This is less tricky than the one on the homework, since we can just make the sequence

(let q~= (x, y) on the line y = x), make the sequence q1~ = (x, y + 1), q2~ =

x, y +1

2

, q3~ =

x, y +

1

3

,...

Q: (THROWN OUT QUESTION FROM ORIGINAL MIDTERM) Prove the following. From Rudin;

Let f: (0,)Rbe any funtion 2x differentiable on its domain.Let M0 = sup {|fx| : x > 0}. Let M1 = sup{|f0x| : x > 0}, and M2 = sup {|f00x| : x > 0}.Show that if M0, M1, M2 are all


4/5

A: Using the following walkthrough:

Part (a): First deal with cases M0 = 0and M2 = 0.

Part (b): Show that for anyx > 0and h > 0, there is some c (x, x + 2h) for which:

f0(x) =f(x + 2h) fx

2h f00(c)

h

Part (c): Conclude that for any x > 0, h > 0,|f0x|6 2M0h M2h

Part (d): Put h = M0

M2

q in (c) to getM16 2 M0M2

.

And the following hint: Use Taylors Theorem.

Pf.(a) If M0 = 0, then f0(x) = 0x > 0, so f0 0, f00 0, so M1 = M2 = 0, so of course 026 4 0 0.If M2 = 0, then f00(x) = 0x > 0, so f0C1 for some constant C1R.So f0(x) = C1x + C2 for some numbers C1 and C2.

If C1 =/ 0, then f

as x

, so M0 = +

.

This means C0 is necessarily0, so that f0 0, so fC2.This says M1 = 0, M2 = C2, so that 0264 0 C2. X(b) Looks like I should use Taylors theorem, on (x, x + 2h). Should try to estimate f(x + 2h)by using f(x),f0(x), f00(c).

Note that we know that fis two times differentiable on(0,)andx0, x0+ 2h (0, +)so c (x, x + 2h)s.t.

f(x + 2h) =f x +f0x

1 (2h) +

f00c

2! (2h)2

(Hoping that if we rearrange this that we get what were supposed to)

f(x + 2h) = f x +f0x

1 (2h) +

f00c

2! (2h)2 = fx + 2hf0x + 2h2f00c

2hf0(x) = f(x + 2h) fx 2h2 f00(c)f0(x) =

f(x + 2h) fx2h

hf00(c).X

(c) Note|f0(x)|6=/ |f(x+2h)|2h +|fx |

2hh|f00c|6 M0

2h+

M0

2h+ hM2 =

M0

h + hM2.

(d) Put h = M0

M2

q ,|f0x|6 M0M2

+ M0M2

= 2 M0M2

, true for allx > 0.

Hence,M1 which is the supremum of all |f0x| for x > 0, is s.t. M162 M0M2

, square both sides to get theresult. X

Q: (Another thrown out problem) Let M , Nbe any metric space. Let f: MNbe any function.(a) Write out condition and open set condition for fto be continuous.(b) Prove open set condition implies condition. (You may use that any open ball is open withoutproving it.)

Hint for b. Pick > 0 and pM. So then M(fx) is open in N.How do you get an open set in M, and what point of Mcan you be sure is in that open set?

4


5/5

A: (a) > 0, pM, > 0, s.t.qM, dM(p, q) < dN(fp, fq) < .[NOT: > 0, > 0,p, qM , dM(p, q) < dN(fp, fq) < . This definition is uniformly continuous, i.e.you can pick same for all p; for normal continuity, you dont.]

Uopen in N, fpreU is open in M.(b) Let > 0. Note that fpre(M(fx))is open in M. This implies that for forx

fpre(M(fx))there exists

> 0 s.t. M(x) fpre(M(fx)).Ifx 0M(x), thend(x, x0) < , and sincex, x0 fpre(M(fx)), thenfx, fx0M(fx), so thatd(fx,fx0) < . X

5

real analysis review

Documents